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Is p > q? (1) p/q > 1 (2) (p - q)/q > (p - q)/p

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Is p > q? (1) p/q > 1 (2) (p - q)/q > (p - q)/p  [#permalink]

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New post 01 Mar 2018, 13:25
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A
B
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E

Difficulty:

  75% (hard)

Question Stats:

52% (02:12) correct 48% (01:56) wrong based on 77 sessions

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Is p > q?

(1) \(\frac{p}{q} > 1\)

(2) \(\frac{(p-q)}{q} > \frac{(p-q)}{p}\)
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Re: Is p > q? (1) p/q > 1 (2) (p - q)/q > (p - q)/p  [#permalink]

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New post 01 Mar 2018, 19:23
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raashiedm wrote:
Is p>q?

(1) \(\frac{p}{q} > 1\)
(2) \(\frac{(p-q)}{q} > \frac{(p-q)}{p}\)


Statement 1: Both \(p\) & \(q\) are either positive or negative. if \(p=2\) & \(q=1\), then we have \(p>q\) & \(\frac{p}{q}>1\), but if \(p=-2\) & \(q=-1\), then \(\frac{p}{q}>1\) but \(p<q\). Hence Insufficient

Statement 2: \(\frac{p-q}{q}-\frac{p-q}{p}>0 =>\frac{(p-q)^2}{pq}>0\)

Now \((p-q)^2>0\), hence \(pq>0\). This implies both \(p\) & \(q\) are either positive or negative. Same as Statement 1. Insufficient

Combining 1 & 2: We know either \(p\) & \(q\) are positive or negative but we cannot determine the value of \(p\) & \(q\). Hence Insufficient

Option E
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Re: Is p > q? (1) p/q > 1 (2) (p - q)/q > (p - q)/p  [#permalink]

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New post 03 Mar 2018, 01:26
raashiedm wrote:
Is p>q?

(1) \(\frac{p}{q} > 1\)
(2) \(\frac{(p-q)}{q} > \frac{(p-q)}{p}\)


Easy E.

Plug in the numbers -3 and -2 for both the statements. Both statements together are not sufficient to answer the question.
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Re: Is p > q? (1) p/q > 1 (2) (p - q)/q > (p - q)/p  [#permalink]

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New post 08 Mar 2018, 02:07
raashiedm wrote:
Is p > q?

(1) \(\frac{p}{q} > 1\)

(2) \(\frac{(p-q)}{q} > \frac{(p-q)}{p}\)


9. Inequalities



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Re: Is p > q? (1) p/q > 1 (2) (p - q)/q > (p - q)/p  [#permalink]

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New post 08 Mar 2018, 07:08
raashiedm wrote:
Is p > q?

(1) \(\frac{p}{q} > 1\)

(2) \(\frac{(p-q)}{q} > \frac{(p-q)}{p}\)


Neither of the statements gives us any information about the sign of p and q. Both the statements are insufficient to answer.

Answer should be E
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Re: Is p > q? (1) p/q > 1 (2) (p - q)/q > (p - q)/p &nbs [#permalink] 08 Mar 2018, 07:08
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